Question

Two particles oscillate in simple harmonic motion along a common straight-line segment of length 0.73 m. Each particle has a period of 3.8 s, but they differ in phase by π/8 rad. (a) How far apart are they 1.2 s after the lagging particle leaves one end of the path? (b) Are they then moving in the same direction, toward each other, or away from each other?

Answer 1

Final Answer:-

A) 0.11 m

B) Same direction

Solution:-

We have two particles which oscillate in simple harmonic motion along common straight-line segment of length A= 0.73 m, so the amplitude of the oscillations for each of the particles is A/2, each particle has a period of T = 3.8 s, but they differ in phase by π/8 rad. Let the phase of the first particle be zero so the phase of the second particle is π/8. So we can write the coordinates of each of the particles as

X1 = A/2 * Cos (w*t)

X2 = A/2 * Cos(w*t + π/8)

now putting the given values,

X1 = 0.73/2 * Cos (2π * 1.2/3.8)

X1 = -0.15 m

X2 = 0.73/2 * Cos (2π * 1.2/3.8 + π/8)

X2 = - 0.26 m

Now, their separation is

∆X = X1 - X2

∆X = -0.15 + 0.26 = 0.11 m

B) Now for directions, we will take the derivative of positions with respect to time.

V1 = dX1/dt

V1 = - (π*A/T) * Sin (2π*t/T)

using known values,

V1 = - 0.55 m/s

Similarly for V2,

V2 = -(π*A/T) * Sin (2π*t/T + π/8)

V2 = - 0.42 m/s

as both V1 and V2 are negative, which means they are moving in the same direction.

* Hope it helps.